What are sign of S from gas moles?

Base the prediction on the change in moles of gas, not total moles; forming a gas from solids gives a large positive S.

Predict the sign of the entropy change for the reaction 2H_2O_2(l) → 2H_2O(l) + O_2(g). [1 mark]

A reaction has H = -100\ kJ mol^-1 and S = +50\ J K^-1mol^-1. Calculate G at 300\ K and state whether it is feasible. [2 marks]

WJEC Eduqas Eduqas 2019-style practice: For the decomposition CaCO_3(s) → CaO(s) + CO_2(g), H = +178\ kJ mol^-1 and S = +161\ J K^-1mol^-1. (a) Calculate G at 298\ K. (b) Determine the minimum temperature at which the reaction becomes feasible.

(a) G = H - T S = 178 - 298 × 1611000 = 178 - 48.0 = +130\ kJ mol^-1 (2). Positive, so not feasible at 298\ K. (b) The reaction becomes feasible when G = 0: T = H S = 1780.161 = 1106\ K (3). Markers reward converting S to kJ, the G calculation, and the feasibility temperature from G = 0.

EnglandChemistrySyllabus dot point

Why do some endothermic reactions happen spontaneously, and how does entropy decide feasibility?

Entropy and its changes, the total entropy change, Gibbs free energy, the condition for feasibility, and the effect of temperature on the feasibility of a reaction.

An Eduqas A-Level Chemistry PI4.2 answer on entropy and entropy changes, Gibbs free energy, the condition for feasibility, and how temperature affects whether a reaction occurs.

Generated by Claude Opus 4.812 min answerUpdated 2026-06-02

Reviewed by: AI editorial process; not yet individually human-reviewed

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What this topic is asking
Entropy and its changes
Gibbs free energy and feasibility
The effect of temperature
Feasibility and reality
Examples in context
Try this

What this topic is asking

Eduqas topic PI4.2 introduces entropy and the thermodynamic condition for a reaction to be feasible. You calculate entropy changes, combine them with enthalpy through Gibbs free energy, apply the feasibility condition $\Delta G \leq 0$ , and explain how temperature can switch a reaction from infeasible to feasible. It explains why some endothermic reactions happen spontaneously.

Entropy and its changes

A reliable guide to the sign of $\Delta S$ is the change in the number of moles of gas: more gas means a positive $\Delta S$ (more disorder); fewer gas molecules means a negative $\Delta S$ .

Gibbs free energy and feasibility

This explains the puzzle: an endothermic reaction ( $\Delta H > 0$ ) can still be feasible if $\Delta S$ is positive and $T$ is high enough that $T\Delta S$ exceeds $\Delta H$ .

The effect of temperature

Finding the feasibility temperature

For a reaction with $\Delta H = +178\ \text{kJ mol}^{-1}$ and $\Delta S = +161\ \text{J K}^{-1}\text{mol}^{-1}$ , find the temperature above which it is feasible.

step 1: Set the feasibility condition

The reaction becomes feasible when $\Delta G = 0$ , i.e. $\Delta H = T\Delta S$ .

step 2: Convert units and rearrange

$\Delta S = 0.161\ \text{kJ K}^{-1}\text{mol}^{-1}$ , so $T = \dfrac{\Delta H}{\Delta S} = \dfrac{178}{0.161}$ .

step 3: Evaluate

$T = 1106\ \text{K}$ . Above this temperature $\Delta G$ is negative and the reaction is feasible; below it, it is not. This is why calcium carbonate must be heated strongly to decompose.

Feasibility and reality

A negative $\Delta G$ means a reaction is thermodynamically feasible, but, as with electrode potentials, it says nothing about rate: a feasible reaction with a high activation energy can be immeasurably slow at room temperature (the combustion of diamond is feasible but does not happen spontaneously).

Examples in context

Example 1. Why ice melts above 0 degrees C. Melting is endothermic but increases entropy (solid to liquid); above the melting point the $T\Delta S$ term outweighs $\Delta H$ , so $\Delta G$ becomes negative and melting is spontaneous.

Example 2. Extracting metals. Reducing a metal oxide with carbon becomes feasible only above a temperature where the entropy gain from producing $\text{CO}$ or $\text{CO}_2$ gas makes $\Delta G$ negative, the thermodynamic basis of the blast furnace.

Try this

Q1. Predict the sign of the entropy change for the reaction $2\text{H}_2\text{O}_2(\text{l}) \rightarrow 2\text{H}_2\text{O}(\text{l}) + \text{O}_2(\text{g})$ . [1 mark]

Cue. Positive, because a gas ( $\text{O}_2$ ) is produced from a liquid, increasing the disorder.

Q2. A reaction has $\Delta H = -100\ \text{kJ mol}^{-1}$ and $\Delta S = +50\ \text{J K}^{-1}\text{mol}^{-1}$ . Calculate $\Delta G$ at $300\ \text{K}$ and state whether it is feasible. [2 marks]

Cue. $\Delta G = -100 - 300 \times 0.050 = -100 - 15 = -115\ \text{kJ mol}^{-1}$ ; negative, so feasible.

Exam-style practice questions

Practice questions written in the style of WJEC Eduqas exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

Eduqas 20195 marksFor the decomposition

\text{CaCO}_3(\text{s}) \rightarrow \text{CaO}(\text{s}) + \text{CO}_2(\text{g})

\Delta H = +178\ \text{kJ mol}^{-1}

and

\Delta S = +161\ \text{J K}^{-1}\text{mol}^{-1}

. (a) Calculate

\Delta G

298\ \text{K}

. (b) Determine the minimum temperature at which the reaction becomes feasible.

Show worked answer →

(a) $\Delta G = \Delta H - T\Delta S = 178 - 298 \times \dfrac{161}{1000} = 178 - 48.0 = +130\ \text{kJ mol}^{-1}$ (2). Positive, so not feasible at $298\ \text{K}$ .

(b) The reaction becomes feasible when $\Delta G = 0$ : $T = \dfrac{\Delta H}{\Delta S} = \dfrac{178}{0.161} = 1106\ \text{K}$ (3).

Markers reward converting $\Delta S$ to $\text{kJ}$ , the $\Delta G$ calculation, and the feasibility temperature from $\Delta G = 0$ .

Eduqas 20214 marks(a) State what is meant by entropy. (b) Predict and explain the sign of the entropy change for

\text{N}_2(\text{g}) + 3\text{H}_2(\text{g}) \rightarrow 2\text{NH}_3(\text{g})

Show worked answer →

(a) Entropy is a measure of the disorder (the number of ways energy and particles can be arranged) of a system; a more disordered system has a higher entropy (1).

(b) The entropy change is negative (1). There are 4 moles of gas on the left and 2 moles of gas on the right, so the number of gas molecules decreases (1), giving fewer ways to arrange the particles and so a decrease in disorder and entropy (1).

Related dot points

Sources & how we know this

WJEC Eduqas GCE A Level Chemistry specification (from 2015) — WJEC Eduqas (2015)